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Math Help - Difficult Series Question

  1. #1
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    Difficult Series Question

    Find the value of b for which

    (summation function; n=0, infinity) e^nb = 1 + e^b + e^2b + e^3b + иии = 9

    How do I go about doing this? Thanks!
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  2. #2
    Super Member TheChaz's Avatar
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    Re: Difficult Series Question

    Well, this is an infinite geometric series with first term = 1 and ratio = e^b.
    There's a formula for that (and maybe even a smartphone app!)...
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  3. #3
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    Re: Difficult Series Question

    Quote Originally Posted by bhaktir View Post
    Find the value of b for which

    (summation function; n=0, infinity) e^nb = 1 + e^b + e^2b + e^3b + иии = 9

    How do I go about doing this? Thanks!
    Hi bhaktir,

    \sum_{n=0}^{\infty}e^{nb}=1 + e^{b} + e^{2b} + e^{3b} +\cdots= 9

    e^{b} + e^{2b} + e^{3b} + \cdots= 8

    e^{b}\left(1 + e^{b} + e^{2b} + e^{3b} +\cdots \right)=8

    e^{b}\sum_{n=0}^{\infty}e^{nb}=8

    9e^{b}=8~;~\mbox{Since, }\sum_{n=0}^{\infty}e^{nb}=9

    e^{b}=\frac{8}{9}

    b=\ln\left(\frac{8}{9}\right)
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  4. #4
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    Re: Difficult Series Question

    Really helpful! Thank you so much, Sudhraka!!
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